Is this 3D curve a circle? The following is a curve in $3$ dimensions:
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/3) \\ z & = &  \cos(\theta - 2\pi/3) \end{eqnarray}$$
Is the curve a circle?
If it is, what about this curve in $4$ dimensions?
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/4) \\
z & = & \cos(\theta - 2\pi/4) \\ w & = & \cos(\theta - 3\pi/4) \end{eqnarray}$$
I don't know if there is something like a circle in $4$-D. If there is, is this curve the $4$-D version of a circle?
P.S.: Is two-dimensional subspace the generalized plane? I want to learn more about this. What should I read?
 A: If you use basic trigonometric identities, you can show the first set of three equations is a piece of this plane: $y-z=x$.
A: Using trigonometric identities, we get
$$\cos(\theta-\frac\pi3) = \cos\theta\cos\frac\pi3+\sin\theta\sin\frac\pi3=\frac12\cos\theta+\frac12\sqrt3\sin\theta,$$
$$\cos(\theta-\frac{2\pi}3) = \cos\theta\cos\frac{2\pi}3+\sin\theta\sin\frac{2\pi}3=-\frac12\cos\theta+\frac12\sqrt3\sin\theta.$$
Thus, if we write $\gamma=(x,y,z)$ we get $\gamma = v_1\cos\theta + v_2\sin\theta$ with $v_1=(1,\frac12,-\frac12)$ and $v2=(0,\frac12\sqrt3,\frac12\sqrt3)$. Thus, we have at least an ellipse. Moreover, it is easy to check that $v1\cdot v2=0$. Furthermore, $v_1^2 = 1 + \frac14+\frac14=\frac32$ and $v_2^2=\frac34+\frac34=\frac32$; thus the vectors are orthogonal and of equal length, and therefore it is indeed a circle.
In the general case, we have $(v_1)_k = \cos \frac{(k-1)\pi}d$ and $(v_2)_k=\sin \frac{(k-1)\pi}d$, where $d$ is the dimension of the vector space ($d=4$ in your second case). Thus, $$v_1\cdot v_2 = \sum_{k=1}^d\cos\frac{(k-1)\pi}d\sin\frac{(k-1)\pi}d = \frac12\sum_{k=1}^d\sin\frac{(k+1)2\pi}{d} = 0$$ and $$v_2^2-v_1^2 = \sum_{k=1}^d\left(\cos^2\frac{(k+1)\pi}d-\sin^2\frac{(k+1)\pi}d\right) = \sum_{k=1}^d\cos\frac{(k+1)2\pi}{d} = 0,$$ The construction thus gives a circle in any dimension.
A: Here is a simple treatment of the $n$-dimensional case, where the point on the curve is $\vec x = (x_0, \ldots, x_{n-1})$ with $x_i = \cos(\theta-\pi i/n)$.

*

*The curve lies on a sphere.
We have $x_i^2 = \cos^2(\theta-\pi i/n) = \frac12+\frac12\cos(2\theta-2\pi i/n)$. So $$\|\vec x\|^2 = \sum x_i^2 = \frac n2 + \frac12 \sum \cos(2\theta-2\pi i/n).$$ The latter term is zero because it is the sum of $n$ equally spaced sinusoids (it is equivalently the $x$-component of the sum of $n$ unit vectors equally spaced along the unit circle, or the real part of the sum of the $n$th roots of $e^{2n\theta\sqrt{-1}}$; in either case, the entire thing is zero by symmetry). So $\|\vec x\|^2$ is a constant, $\frac n2$, independent of $\theta$.


*The curve lies on a two-dimensional subspace.
We have $x_i = \cos(\theta-\pi i/n) = a_i\cos\theta + b_i\sin\theta$ for some fixed $a_i$ and $b_i$ independent of $\theta$. Then $\vec x = \vec a\cos\theta + \vec b\sin\theta$. So $\vec x$ lies on the two-dimensional subspace spanned by $\vec a$ and $\vec b$.
Thus, $\vec x$ lies on the intersection of a sphere in $n$ dimensions and a two-dimensional subspace, i.e. a sphere in two dimensions, also known as a circle.
A: $\cos3\theta=cos3(\theta)$
$\cos3(\theta-\frac{\pi}{3})=cos(3\theta-\pi)=-\cos3\theta$
As $-y=-\cos(\theta-\frac{\pi}{3})=cos(\theta+\frac{2\pi}{3})$,
$\cos3(\theta+\frac{\pi}{3})=\cos(2\pi+3\theta)=\cos3\theta$
$\cos3(\theta-\frac{2\pi}{3})=cos(3\theta-2\pi)=\cos3\theta$
Now, $\cos3\theta=4cos^3\theta-3\cos\theta$
If $\cos3\theta=a$ and $\cos\theta=t$
So, $x,-y,z$ are the roots of  $4t^3-3t-a=0$
$=>x+(-y)+z=0$
$=>x(-y)+(-y)z+zx=\frac{3}{4}$
$=>x^2+y^2+z^2=(x+(-y)+z)^2-2(x(-y)+(-y)z+zx)=0+2\frac{3}{4}$
$=>x^2+y^2+z^2=\frac{3}{2}$
Observe that $(x,y,z)$  satisfy a general plane equation $Ax+By+CZ+D=0$   where $A,B,C,D $ constants,  not all zeros.  
Also, satisfies the equation of the general circle in 3-D, $(x-a)^2+(y-b)^2+(z-c)^2=d^2 $.  $a=b=c=0, d^2=\frac{3}{2}$
In case of $x,y,z,w$,
$z=\cos(\theta-\frac{2\pi}{4})=\sin\theta$,
$\sqrt2 y=\cos\theta+\sin\theta$,
$\sqrt2 w=-\cos\theta+\sin\theta$
So,$x^2+z^2=1$  and $w^2+y^2=1$
$x^2+y^2+z^2+w^2=2$
Now $\sqrt2 (y+w)=2\sin\theta=2z=>\sqrt 2z=y+w$
Similarly, $y-w=\sqrt 2x$ 
Observe that   $(x,y,z,w)$  satisfies two general plane equations  $Ax+By+CZ+Dw=E$  where $A,B,C,D,E $ constants, not all zeros. 
Also, satisfies the equation of the general circle in  4-D, $(x-a)^2+(y-b)^2+(z-c)^2+(w-d)^2=e^2 $ .  $a=b=c=d=0, e^2=2$
Again, we know $\cos nx=$Real part of $(\cos x+i\sin x)^n=(\cos x)^n+^nC_2(\cos x)^{n-2}(\sin x)^2+^nC_4(\cos x)^{n-4}(\sin x)^4+...$
Observe there is no term containing  $=(\cos x)^{n-1}$
As $\cos n(2x-\frac{2r_i\pi}{n})=\cos(2nx-2r_i\pi)=\cos 2nx=C(say)$ 
So, $\cos (2x-\frac{2r_i\pi}{n})=R_i$(say), where all $r_i$s are distinct integers with   $0 ≤r_i< n$ are the roots of the equation 
$2^{n-1}y^n+C_1y^{n-2}+...-C=0$
So, $\sum R_i=0$ as the coefficient of $y^{n-1}$ is 0.
If $x_i=\cos (x-\frac{r_i\pi}{n})=>R_i=2(x_i)^2-1$ 
So, $\sum (2(x_i)^2-1)=0 =>\sum (x_i)^2=\frac{n}{2}$
This is another way of generalization("The curve lies on a sphere") already achieved by Rahul Narain. 
A: For at least the three-dimensional case, here's a more mechanical method (i.e. less enlightening than Rahul's nice answer) to verify if your curve is a circle:
We try to evaluate the curvature $\kappa$ and torsion $\tau$ of the given curve. By the fundamental theorem of space curves, a space curve is uniquely determined (up to rigid motions) by $\kappa(s)$ and $\tau(s)$; if, in addition, $\tau(s)=0$ (i.e. the curve is flat) and $\kappa(s)$ is a constant greater than zero, then we know that the space curve is indeed a circle.
Using formula 26 here for the curvature, we have
$$\begin{align*}
\kappa&=\frac{\|\mathbf r^\prime\times \mathbf r^{\prime\prime}\|}{\|\mathbf r^\prime\|^3}\\
&=\frac{\left\|\langle-\sin\,\theta,\cos\left(\theta+\frac{\pi}{6}\right),\cos\left(\theta-\frac{\pi}{6}\right)\rangle\times\langle-\cos\,\theta,-\sin\left(\theta+\frac{\pi}{6}\right),\sin\left(\frac{\pi}{6}-\theta\right)\rangle\right\|}{\left\|\langle-\sin\,\theta,\cos\left(\theta+\frac{\pi}{6}\right),\cos\left(\theta-\frac{\pi}{6}\right)\rangle\right\|^3}\\
&=\sqrt\frac23
\end{align*}$$
Using formula 3 here for the torsion, we have
$$\begin{align*}
\tau&=\frac{\mathbf r^\prime\times \mathbf r^{\prime\prime}\cdot\mathbf r^{\prime\prime\prime}}{\kappa^2}\\
&=\frac1{2/3}\begin{vmatrix}-\sin\,\theta&\cos\left(\theta+\frac{\pi}{6}\right)&\cos\left(\theta-\frac{\pi}{6}\right)\\-\cos\,\theta&-\sin\left(\theta+\frac{\pi}{6}\right)&\sin\left(\frac{\pi}{6}-\theta\right)\\\sin\,\theta&-\cos\left(\theta+\frac{\pi}{6}\right)&-\cos\left(\theta-\frac{\pi}{6}\right)\end{vmatrix}\\
&=0
\end{align*}$$
(Note that the determinant is easily seen to be zero, since the third row is a multiple of the first.)

Since $\tau=0$, the curve is flat; in addition, since $\kappa=\sqrt{2/3}$, we find that our space curve is a circle with radius $1/\kappa=\sqrt{3/2}$.
A: It is a circle. If you look at the distance from the origin
$$ r = \sqrt{ \cos^2(\theta) + \cos^2(\theta+\frac{\pi}{3}) + \cos^2(\theta+\frac{2\pi}{3}) } $$
which simplifies to $r = \frac{\sqrt{6}}{2} $. 
With the 4-dimensional case the distance simplifies to $r=\sqrt{2}$. 
I wonder how to prove the general case of
$$ r^2 = \sum_{i=1}^N \left[ \cos^2\left( \theta + \frac{i-1}{N} \pi \right) \right] = \frac{N}{2} $$
